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We had to compute the integral $$\int \frac{3}{x^3-1} dx$$ I am just starting to learn about integration, and I have no idea how to begin. Can you please help me solve this? Even a hint would be very helpful. Thank you in advance!

P.S. This question is not really a duplicate because the numerator is 3 not 1, so it is harder. If it were 1, I could do it easily.

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    $\begingroup$ Do you know Partial fraction decomposition? $\endgroup$
    – Sorfosh
    May 8 '18 at 2:40
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    $\begingroup$ Are you familiar with partial fractions? $\endgroup$
    – User1234
    May 8 '18 at 2:40
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    $\begingroup$ If so, you can break the denominator into two parts:$(x-1)$ and $(x^2+1+x)$. $\endgroup$
    – User1234
    May 8 '18 at 2:41
  • $\begingroup$ Did you really have to post it here to get your "brother" to do it for you? $\endgroup$ May 8 '18 at 4:11
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    $\begingroup$ what do you mean? $\endgroup$
    – user547340
    May 8 '18 at 5:43
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Using partial fraction decomposition, you get (as Macrophange said) $$\int (\frac {1}{x-1}-\frac {x+2}{x^2+x+1}) dx$$ The left is straightforward: $$\int (\frac {1}{x-1}) dx =\log |x-1|+C$$ The right is more complicated, because I cannot factor the demonstrator, and I tried the Chain Rule, which unfortunately did not work. (I got $u=x^2+x+1$ gives $du=(2x+1) dx$, which is not the denominator) However, I found that we can force this substitution to match partially, by $\frac {x+2}{x^2+x+1}=\frac{1}{2}\cdot\frac {2x+1}{x^2+x+1}+\frac{3}{2}\cdot\frac {1}{x^2+x+1}$. Now, on the first summand, applying the chain rule gives us $$\int (\frac{1}{2}\cdot\frac {2x+1}{x^2+x+1}) dx=\int (\frac{1}{2}\frac {du}{u})=\frac {1}{2} \log|u|+C=\frac {1}{2} \log|x^2+x+1|+C$$ The second term that is left, $\frac{3}{2}\cdot\frac {1}{x^2+x+1}$, looks suspiciously like the arctangent, but with an extra linear term. We can make it closer by completing the square in the denominator, which results in $$\frac {1}{(x+\frac {1}{2})^2+\frac {3}{4}}$$. We evaluate the antiderivative of this via trig substitution, and I don't think you have learned it yet. If you do, you can do the rest by yourself. The answer is a true monster: $$\log |x-1|-\frac{1}{2} \log|x^2+x+1|-\sqrt {3}\tan^{-1}(\frac {2}{\sqrt 3} (x+\frac {1}{2})+C$$ Whew!

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  • $\begingroup$ Took me some time to work it out. $\endgroup$
    – KingLogic
    May 8 '18 at 3:05
  • $\begingroup$ Thank you, but can you please work out the trig substitution? $\endgroup$
    – user547340
    May 8 '18 at 4:07
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Hint: Use partial fraction decomposition. The gist is that you have to factor the denominator into linear terms$(ax+b)$ or irreducible quadratic terms$(ax^2+bx+c)$. Next, you expand the fraction into several simpler fractions to integrate. Try it yourself to find some patterns, :)

$$\int \frac{3}{x^3-1} dx=\int\frac{3}{(x-1)(x^2+x+1)}=\int(-\frac{x+2}{x^2+x+1}+\frac{1}{x-1})dx$$

And then you can proceed to find the indefinite integral of each part in the integrand(Hint: you should get $\ln(\dots)$ and $\arctan(\dots)$). You can derive the arctan part with completing squares and trigonometric substitution.

Please do ask if you have more questions, hope it can help you!

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    $\begingroup$ @imranfat Makes sense, so I added some description about PFD. $\endgroup$
    – Macrophage
    May 8 '18 at 2:53
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    $\begingroup$ @FamousMichaelWang It looks all right to me. Good job! $\endgroup$
    – Macrophage
    May 8 '18 at 3:57
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    $\begingroup$ @Macrophange Thank you very much! $\endgroup$
    – KingLogic
    May 8 '18 at 3:59
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    $\begingroup$ I attributed to you in my answer (partial fraction decomposition) $\endgroup$
    – KingLogic
    May 8 '18 at 3:59
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    $\begingroup$ Thanks dude. But it's you who completed the rest of the integration. Cheers! $\endgroup$
    – Macrophage
    May 8 '18 at 4:00